2003 Trig Contest

[prog_tom]

WELCOME TO TOM'S 2003 TRIG COMPETITION

(1). Determine the solution to the inequality |x3 - 5x + 3| - 4 < 0

(2). A window had the shape of a rectangle with a semicircle mounted on the top. Let x represent the diameter of the semicircle. The rectangular portion of the rectangle is 5ft taller than its wide. Give the Area of the window as a function of x.

(3). The halflife of tritium is 12.4 years. If the initial amount of tritium present is 0.42 kg, approx how much tritium will remain after 45 months?

[sql_lall]

#2: Area=area rect + area circ/2.
Area rect = length * height = x(x+5)
Area circ/2 = (Pi * r^2)/2 = Pi*x^2/4/2 = (Pi*x^2)/8
=> Formula = x(x+5) + (Pi*x^2)/8
*This can be mixed around to be a generic quadratic if needed.

#3: The function can be written thus:
W_t = W₀ * 2^(-t/12.4)

W_t = weight after t years
W₀ = origional weight
t = time in years

t = 45/12 = 15/4
W₀ = 0.42
=> W_15/4 = 0.42 * 2^(-15/49.6)
=. W_15/4 = 0.3406 (4 s.f.)

[TheAlchemist]

hey prog_tom,
i think theres a problem with the equation in the first question:

|x^3 - 5x + 3| -4 < 0 ---assertion, test with x =3

|(3^3) - (5*3) + 3| - 4 = 11 which is > 0

i tried it with other values too but it only worked with x =1 and
x = 2.

sql_lall has done justice to the others!

[prog_tom]

Originally posted by sql_lall
#2: Area=area rect + area circ/2.
Area rect = length * height = x(x+5)
Area circ/2 = (Pi * r^2)/2 = Pi*x^2/4/2 = (Pi*x^2)/8
=> Formula = x(x+5) + (Pi*x^2)/8
*This can be mixed around to be a generic quadratic if needed.

#3: The function can be written thus:
W_t = W₀ * 2^(-t/12.4)

W_t = weight after t years
W₀ = origional weight
t = time in years

t = 45/12 = 15/4
W₀ = 0.42
=> W_15/4 = 0.42 * 2^(-15/49.6)
=. W_15/4 = 0.3406 (4 s.f.)

What unit is your answer for 3 in?

[sql_lall]

Good point:

kg * integer = kg, so my answer was in Kilograms.